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A bicycle tire has a pressure of 7.00×105N/m2at a temperature of 18.00Cand contains2.00L of gas. What will its pressure be if you let out an amount of air that has a volume of100cm3 at atmospheric pressure? Assume tire temperature and volume remain constant.

Short Answer

Expert verified

The pressure of the gas becomes6.93×105N/m.

Step by step solution

01

State the ideal gas law.

According to the ideal gas law,
PV=NkT

Here, is the pressure of the gas, is the volume occupied by the gas, is the number of molecules in the gas, is the Boltzmann constant, and is the absolute temperature.

02

Calculate the number of molecules present in the gas initially.

Let be the initial pressure, be the number of molecules present initially, be the final pressure, and be the number of molecules at final pressure. Since volume and temperature are constant, the initial and final temperatures can be denoted as Likewise, initial and final volume can be represented as .

P0=7.00×105N/m2TK=18.00C+273.15T=291.15Kk=1.38×10-23J/K

role="math" localid="1654860621975" V=2.00L1000L/m3=2×10-3m3

Substitute the values in the equation

7.00×105N/m2×2.00×10-3m3=Ni×1.38×10-23J/K×291.15KNi=7.00×105N/m2×2.00×10-3m31.38×10-23J/K×291.15KNi=3.48×1023

Therefore, the number of molecules present in the gas initially is 3.48×1023.

03

Calculate the number of molecules removed from the tire.

Let the volume of the air removed from the air be V1and its pressure will be equal to the atmospheric pressure. It can be denoted as P1. The number of molecules removed from the tire can be denoted as N1.

P1=1.013×105N/m2V1=100cm3106cm3/m3=100×10-6m3

Substitute these values in the ideal gas equation.

1.013×105N/m2×100×10-6m3=N1x1.38×10-23J/K×291.15KNi=1.013×105N/m2×100×10-6m31.38×10-23J/K×291.15KNi=2.52×1021

Therefore, the number of molecules removed from the tire is2.52×1021 .

04

Calculate the final pressure.

Since volume and temperature are constant, the unknown quantity that remains to calculate the pressure is the number of molecules. The number of molecules present at this stage is the difference between the number of molecules present initially and the number of molecules removed from the tire.

Nf=N0-Ni=3.48×1023-2.52×1021=3.45×1023

Substitute these values in the ideal gas equation.

Pf×2×10-3m3=3.45×1023×1.38×10-23J/K×291.15KPf=3.45×1023×1.38×10-23J/K×291.15K2×10-3m3Pf=6.93×105N/m2

Therefore, the final pressure is 6.93×105N/m2.

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